Question

If $\theta$ is the angle between the line
$\overrightarrow{r}=2i+j-k+(i+j+k)t$ and the plane
$\overrightarrow{r}\cdot (3i-4j+5k)=q$, then

• A. $\cos \theta =\frac{2\sqrt{6}}{15}$
• B. $\sin \theta =\frac{2\sqrt{6}}{15}$
• C. $\sin \theta =-\frac{11\sqrt{7}}{70}$
• D. $\cos \theta =-\frac{11\sqrt{7}}{70}$

Answer 1

The correct option is B $\sin \theta =\frac{2\sqrt{6}}{15}$

$\theta$ is angle b/w $\underset{\gamma }{\overset{}{\to }}=2\hat{i}+j+k+(i+j+k)t$ and $\to .(3\hat{i}-4\hat{j}+5k)=q$

Angle b/w line and plane is given by

$\sin \theta =\frac{4_1{a}_2+b_1{b}_2+c_1{c}_2}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}$

Where $(a_1,b_1,c_1)$ and $(a_2,b_2,c_2)$ are direction ratios of line and plane Respectively so here

$a_1,b_1,c_1)=(1,1,1)$ and $(a_2,b_2,c_2)=(3,-4,5)$

So $\sin \theta =\frac{3-4+5}{\sqrt{1+1+1}\sqrt{9+16+25}}$

$\frac{4}{\sqrt{3}\sqrt{50}}=\frac{4}{\sqrt{3}5\sqrt{2}}=\frac{4}{\sqrt{6.5}}\times \frac{\sqrt{6}}{\sqrt{6}}=\frac{2\sqrt{6}}{5.3}=\frac{2\sqrt{6}}{15}$

so here $\sin \theta =\frac{2\sqrt{6}}{15}\Rightarrow \theta =\sin \frac{2\sqrt{6}}{15}$